XII. Second Order Linear Differential Equations
12.1 Homogeneous Equations
A differential equation is an equation involving variables x, y, y 0 , y 00 , . . .. A solution is a function
f (x) defined in an interval I such that the substitution y = f (x), y 0 = f 0 (x), y 00 = f 00 (x), . . .
becomes an identity for x in the interval I. The differential equation is said to be linear if it
is linear in the variables y, y 0 , y 00 , . . .. We first encountered solutions of differential equations in
section 4 of Chapter 3, in the context of equations of the form f (y)dy = g(x)dx). In Chapter 6,
section 3 we discussed the solution of first order linear equations (those in which only the variables
x, y, y 0 appear). in this chapter we turn to second order linear equations with constant coefficients
(those allowing also a second derivative). The general form of such an equation is
(12.1)
y 00 + ay 0 + by = g(x) ,
where a and b are constants, and g(x) is a differentiable function of x. In section 6.4, we saw that a
first order equation has a one-parameter family of solutions, and that the specification of an initial
condition y(x0 ) = y0 uniquely determines a solution. In the case of second order equations, the
basic theorem is this.
Theorem 12.1. Given x0 in the domain of the differentiable function g, and numbers y0 , y00 ,
there is a unique function f (x) which solves the differential equation (12.1) and satisfies the initial
conditions f (x0 ) = y0 , f 0 (x0 ) = y00 .
In this section we shall see how to completely solve equation (12.1) when the function on the right
hand side is zero:
(12.2)
y 00 + ay 0 + by = 0 .
This is called the homogeneous equation. An important first step is to notice that if f (x) and
g(x) are two solutions, then so is the sum; in fact, so is any linear combination Af (x) + Bg(x).
Thus, once we know two solutions (they must be independent in the sense that one isn’t a constant
multiple of the other) we can solve the initial value problem in theorem 12.1 by solving for A and
B.
Example 12.1. Solve.
y 00 + y = 0 ,
y(0) = 4 ,
y 0 (0) = −1 .
Now, we know that cos x and sin x are solutions of the equation, so we try a solution of the
form y(x) = A cos x + B sin x. Evaluating at x = 0, we find that A = 4. Differentiate, getting
y 0 (x) = −A sin x + B cos x, and evaluating at x = 0, we find B = −1. Thus the solution is
y(x) = 4 cos x − sin x.
The reason the answer worked out so easily is that y1 = cos x is the solution with the particular
initial values y1 (0) = 1, y10 (0) = 0 and y1 = sin x is the solution with y1 (0) = 0. y10 (0) = 1 . Then
the solution with initial values y(0) and y 0 (0) is
(12.3)
Example 12.2. Solve
y(x) = y(0) cos x + y 0 (0) sin x
y 00 − y = 0 , with given initial values
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y(0) , y 0 (0) .
Now ex and e−x are solutions of this differential equation, so the general solution is a linear
combination of these. But we won’t have as easy a time finding a solution like (11.3), since these
functions do not have the initial values 1, 0; 0, 1 respectively. However if we introduce the functions
cosh x =
1 x
(e + e−x )
2
sinh x =
1 x
(e − e−x )
2
these do have the right initial values:
cosh 0 = 1 ,
(12.4)
d
(cosh x) = sinh x ,
dx
sinh 0 = 0
d
(sinh x) = cosh x
dx
so (cosh)0 (0) = 0, (sinh)0 (0) = 1. Thus, the solution to our problem is
y(x) = y(0) cosh x + y 0 (0) sinh x .
This particular differential equation comes up so often that it is important to remember these
functions, cosh x, sinh x, called the hyperbolic functions and their basic properties: equation
(12.4) and
(12.5)
cosh2 x − sinh2 x = 1 .
Because of (12.5) these functions parametrize the standard hyperbola (and it is for this reason that
they are called hyperbolic functions).
Using these examples as a guide, we return to the general second order equation.
Proposition 12.2. Let r be a root of the equation
(12.6)
r2 + ar + b = 0 .
Then erx is a solution to the homogeneous equation:
(12.7)
y 00 + ay 0 + by = 0 .
Equation (12.6) is called the auxiliary equation of the differential equation (12.7). To verify the
proposition, let y = erx so that y 0 = rerx , y 00 = r2 erx . Substituting into equation (12.7):
r2 erx + arerx + berx = erx (r2 + ar + b) ,
and this is zero if and only if r is a root of the auxiliary equation.
Now unfortunately a quadratic equation does not necessarily always have two real roots, so we
have to examine the cases separately.
Case of two real roots. If the discriminant a2 − 4b > 0, then there are two real roots, and it is
straightforward to find the solution of the corresponding initial value problem.
Example 12.3. Solve :
y 00 + 6y 0 + 5y = 0 ,
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y(0) = 4 ,
y 0 (0) = −1 .
The auxiliary equation, r2 + 6r + 5 = 0 has the roots r = −1, −5, so e−x and e−5x are solutions.
The general solution is
y = Ae−x + Be−5x
with derivative
y 0 = −Ae−x − 5Be−5x .
Evaluating at x = 0, we have 4 = A + B, −1 = −A − 5B. Solving this pair of equations, we get
A = 19/4 and B = −3/4, so our solution is
y=
19 −x 3 −5x
e − e
4
4
Example 12.4. A function x = x(t) satisfies the differential equation
x00 − 2x0 − 15x = 0 .
Under what conditions on the values of x and x0 at t = 0 will this function decay to 0 as t → ∞?
The auxiliary equation r2 − 2r − 15 has the roots r = −3, 5. Thus the general solution is
x(t) = Ae−3t + Be5t . This will decay at infinity only if B = 0. Now, evaluating x and x0 at 0 gives
us the equations
x(0) = A + B ,
x0 (0) = −3A − 5B .
Setting B = 0, the condition becomes x0 (0) + 3x(0) = 0.
Case of complex roots. If the discriminant a2 −4b < 0, then the roots are two complex conjugate
numbers α + iβ, α − iβ.
Let’s look again at the case y 00 + y = 0. Then the roots of r2 + 1 = 0 are ±i, and we’d like to say
that the solutions are the functions eix , e−ix . This does work, and all the algebra in the case of
real roots works just as well in this case, once we have given these expressions meaning. First of
all, remember equation (12.3): the general solution of y 00 + y = 0 is
(12.8)
y(x) = y(0) cos x + y 0 (0) sin x
If y(x) = eix is to represent a solution of this differential equation, we have y(0) = e0 = 1, and
y 0 (0) = ie0 = i, so we must have
(12.9)
eix = cos x + i sin x
Notice that if we differentiate this expression, we get
− sin x + i cos x = i(cos x + i sin x) ,
so this expression is consistent with the differentiation rule for the exponential:
d ix
e = ieix .
dx
In fact, defining the complex exponential by (12.9) is consistent with all the rules of exponentials.
In particular (recall problem 10 of section 9.5), if we substitute the Maclaurin series for all the
functions in (12.9) we get an identity:
∞
∞
∞
X
X
X
(ix)n
x2n
x2n+1
=
(−1)n
+i
(−1)n
.
n!
(2n)!
(2n + 1)!
n=0
n=0
n=0
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Proposition 12.3. For a complex number α + iβ if we define the exponential function as
e(α+iβ)x = eαx eiβx = eαx (cos(βx) + i sin(βx)) ,
then all the usual laws of exponents carry through.
Now, of course, we are interested only in real-valued functions. What we have shown is that if α±iβ
are the roots of the equation r2 + ar + b = 0, then the functions e(α±iβ)x solve the differential
equation y 00 + ay 0 + b = 0. But then the real and imaginary parts of this function satisfy the
equation as well, which gives us the desired two real-valued solutions.
Proposition 12.4 . If the auxiliary equation for the differential equation
y 00 + ay 0 + b = 0
has the complex roots α ± iβ, then every solution of the differential equation is of the form
(12.10)
Aeαx cos(βx) + Beαx sin(βx) = eαx (A cos(βx) + B sin(βx))
In solving initial value problems, we can work with the complex solutions or solutions of the form
(12.10); usually the latter is more convenient.
Example 12.5. Find the general solution x = x(t) of
x00 + a2 x = 0 .
Since the roots of the auxiliary equation r2 = −a2 = 0 are ±ia, the general solution is
(12.11)
x(t) = A cos at + B sin at
It is easy to see what this function looks like by defining
p
C = A2 + B 2 , γ = arctan(B/A)
Then (12.11) becomes
x(t) = C(cos γ cos at + sin γ sin at = C cos(at − γ) = C cos a(t − γ/a)) .
Thus the graph of x = x(t) is a simple cosine curve of amplitude C, and period 2π/a, shifted to
the right by the phase γ. (See figure 12.1).
Figure 12.1
2
15
1
05
1
0
05 0
05
1
05
15
2
231
1
15
2
Example 12.6. Find the solution y = y(x) of y 00 + 2y 0 + 5y = 0, with initial values y(0) =
2, y 0 (0) = −1.
The auxiliary equation r2 + 2r + 5 = 0 has the solutions r = −1 ± 2i. Thus the general solution
is y = e−x (A cos(2x) + B sin(2x)). To solve for A and B using the initial values we must first
differentiate y:
y 0 = −e−x (A cos(2x) + B sin(2x)) + e−x (−2A sin(2x) + 2B cos(2x)) .
Substituting the initial values gives the equations A = 2, −A + 2B = −1, which has the solutions
A = 2, B = 1/2. The answer thus is
y = e−x (2 cos(2x) +
1
sin(2x)) .
2
Case of a double root. If the discriminant a2 − 4b = 0, then the auxiliary equation has one root
r, which gives us only one solution erx of the differential equation. We find another solution by
the technique of variation of parameters. We try y = uerx , where u is a new unknown function.
Now, since the auxiliary equation has only the single root r, we have a = −2r and b = r2 , so the
differential equation can be written as:
y 00 − 2ry 0 + r2 y = 0 .
Substituting y = uerx in the left hand side, we get
y 00 − 2ry 0 + r2 y = erx [(u00 + 2u0 r + ur2 ) − 2r(u0 + ur) + r2 u] = erx u00 .
For this to be zero, we must have u00 = 0, so that u = Ax + B.
Proposition 12.5 . If the auxiliary equation for the differential equation
y 00 + ay 0 + b = 0
has only the root r, then every solution is of the form
(Ax + B)erx
Example 12.7. Find the solution of y 00 − 4y 0 + 4y = 0 with initial values y(0) = 2, y 0 (0) = −1.
The auxiliary equation has just the root r = 2. The general solution is y = (Ax + B)e2x , with
derivative y 0 = 2(Ax + B)e2x + Ae2x. Substituting the initial conditions gives the equations
2=B
− 1 = 2B + A .
Thus A = −5, B = 2 and the answer is
y = (−5x + 2)e2x .
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Problems 12.1
1. Solve y 00 − 5y = 0 with the initial values y(0) = 1, y 0 (0) = −1.
2. Solve y 00 + 5y = 0 with the initial values y(0) = 1, y 0 (0) = −1.
3. Solve y 00 − 5y 0 + 6y = 0 with the initial values y(0) = 1, y 0 (0) = −1.
4. Solve y 00 + 4y 0 + 5y = 0 with the initial values y(0) = 1, y 0 (0) = −1.
5. Solve y 00 − y 0 = 0 with the initial values y(2) = 1, y 0 (2) = 2.
6. Solve y 00 + 2y 0 + y = 0 with the initial values y(−1) = 1, y(1) = 1.
12.2 Behavior of the Solutions
When these differential equations come up in applications, one usually wants to have some idea
of the long term behavior of the solution. Since the roots of the auxiliary equation determine the
solutions, they also determine their behavior. Here we summarize the results.
Both roots positive. Except for the identically zero solution, all solutions grow exponentially.
See figure 12.2.
Both roots negative. Except for the identically zero solution, all solutions decay exponentially.
See figure 12.3.
Figure 12.3
Figure 12.2
5
7
7
5
4
3
2
1
0
−1 0
4
3
2
1
02
04
06
08
1
0
0
1
2
3
4
5
7
A negative and a positive root. All solutions grow exponentially, except for the multiples of
the exponential with the negative root.
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Both roots imaginary. In this case the equation is
y 00 + ω 2 y = 0 ,
As we saw in example 12.5, the general solution can be written as
or y(x) = C cos(ωx − γ) ,
y(x) = A cos ωx) + B sin(ωx)
an oscillation of period 2π/ω, amplitude C and phase γ (see figure 12.1).
Figure 12.1
2
15
1
05
1
0
05 0
05
1
05
15
1
2
15
2
Complex roots. In this case the roots are of the form α ± iβ and the general solution can be
written in the form (following the previous discussion)
y(x) = Ceαx cos(ωx − γ) .
Thus if α > 0, this gives an oscillation with exponentially increasing amplitude (figure 12.4), and
if α < 0, this gives an exponentially damped oscillation (figure 12.5).
Figure 12.5
Figure 12.4
500
1
300
05
100
0
100 0
0
1
2
3
4
5
6
0
300
05
500
234
1
1
2
3
4
5
6
Problems 12.2
For each of the following differential equations, find the general solution in terms of the initial
values y(0), y 0 (0). Give conditions, if any, on the initial conditions for the solution to be bounded.
1.
y 00 − 6y 0 + 9y = 0 .
2.
00
− 6y 0 + 5y = 0 .
3.
y 00 − y 0 + 6y = 0 .
4.
y 00 + 4y 0 + 13y = 0 .
5.
y 00 − 4y 0 + 13y = 0 .
6.
y 00 + 9y = 0 .
7.
y 00 − 9y = 0 .
8. Find the solution of problem 3 with initial values y(0) = 3, y 0 (0) = −2.
9. Find the solution of problem 4 with initial values y(0) = 0, y 0 (0) = 1.
12.3 Applications
Springs
Suppose we place a mass m on the end of a vertically hanging spring, and then set it in motion;
how can we describe the subsequent motion? As we have seen in chapter 5, section 4, the spring is
subject to a restoring force proportional to its displacement from equilibrium. By Newton’s second
law of motion, this force is ma, where a is the acceleration of the mass m. Letting x represent the
downward displacement from equilibrium, we have a = x00 , and if the spring constant is k, this
gives us the equation
(12.12)
Letting ω =
mx00 = −kx ,
or x00 +
k
x=0
m
p
k/m, this has the solution (see example 12.5)
x(t) = C cos(ωt − γ)
where C and γ are to be determined by the initial data.
We have to be a little careful about units. In the metric system, when m is measured in kilograms
and x in meters, then force is measured in newtons, and the units for the spring constant k are
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newtons/meter. On a smaller scale, m is in grams, x in centimeters, force in dynes, and k in
dynes/meter. However, in the British system it is customary to refer to the weight w of the object
(in pounds), rather than its mass. Then m = w/g, where g = 32 ft/sec2 is the acceleration due to
gravity. Finally, in the British system, the spring constant is given in lbs/foot.
Example 12.8. Suppose a mass of 10 g hangs from a spring with spring constant k = .4 dynes/cm.
If the spring is extended an additional 8 cm. and then released, give the equation of subsequent
motion.
p
0
.4/10 =
The
initial
conditions
are
that
when
t
=
0,
x(0)
=
8,
x
(0)
=
0.
We
also
have
ω
=
√
.04 = .2. Thus the solution has the form
x(t) = C cos(.2t + γ)
We get, from the initial conditions
0 = −.2C sin(−γ)
8 = C cos(−γ) ,
so γ = 0 and C = 8, and the equation of motion is
x(t) = 8 cos(.2t)
We could have concluded this more quickly, by observing that the initial conditions tell us that
x = 8 when the velocity is 0, so the maximum extension (the amplitude) has to be 8.
Example 12.9. Suppose that we come upon the above configuration already in motion, and when
we make our observation (at time t = 0), the mass is displaced 12 cm downward and is traveling
downward at a velocity of 1 cm/sec. Find the equation of motion.
Again, the equation has the general form
x(t) = C cos(.2t − γ)
and the initial conditions give
1 = −.2C sin(−γ) .
12 = C cos(−γ) ,
We solve for C and γ as follows. The equations are
C cos(−γ) = 12 .
C sin(−γ) = −5 .
Adding the squares of both equations gives us C 2 = 122 + 52 = 169, so C = 13, and dividing one
equation by the other gives
tan(−γ) = −5/12‘,
so that γ = .126π
and the equation of motion is
x(t) = 13 cos(.2t − .126π) .
Example 12.10. If a 16 lb. object is hung from a spring with spring constant k=9 lbs/foot and
then is given an initial velocity of 24 ft/sec, what is the maximum extent of the spring?
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Here m = 16/32 and k = 9, so we have the spring equation
1 00
x + 9x = 0
2
√
√
so√x = A cos(3 2t) + B sin(3 2t). The initial conditions x(0) = 0, x0 (0) = 24 lead to A = 0, B =
8 2. The solution thus is
√
√
x(t) = 8 2 sin(2 2t)
√
whose maximum value is 8 2 feet.
Now, let us return to our spring with spring constant k and mass m, and suppose that it is inserted
in a viscous fluid which imparts a retarding force proportional to the velocity of the mass. Letting
q > 0 be the constant in this proportion, we see that equation (12.12) is replaced by
mx00 = −kx − qx0 ,
(12.13)
where ω =
or x00 + νx0 + ω 2 x = 0
p
k/m and ν = q/m. The roots of this equation are
√
−ν ± ν 2 − 4ω 2
r=
.
2
If ν > 2ω, then the roots are both real and negative, and there is exponential decay with no
oscillation. if ν < 2ω, then we have complex roots, so there are oscillations. But the real part of
the roots is −ν/2 < 0, so the oscillations are exponentially damped. Thus, if we want to have a
good damping effect (as for example in a shock absorber) we should be sure that ν is sufficiently
large; that is, that the fluid is very viscous.
Example 12.11. A system consisting of a spring in a viscous fluid is installed so as to absorb the
shock on a 100 kg mass. The spring constant is k=2500 and the constant of viscosity is q=600.
Suppose a shock is sustained when the system is in equilibrium imparting an instantaneous velocity
of 100 cm/sec. How long will it take for the amplitude of the oscillation to be reduced to 1 cm?
The basic differential equation is 100x00 + 600x0 + 2500x = 0. The roots of the auxiliary equation
are r = −3 ± 4i, so the general solution is
x(t) = e−3t (A cos 4t + B sin 4t)
Now at t = 0, x = 0, x0 = 100. Solving for A and B with those initial conditions, we find
x(t) = 25e−3t sin 4t. Now the maximum amplitudes of this damped vibration occur at the values
t = kπ/8, for k an odd integer. Here is a table of the first few values:
t
A
.3927
7.697
1.1781
.729
1.9634
.069
Thus the first maximum amplitude occurs at .3927 seconds and is 7.697 cm, but by 1.1781 the
maximum amplitude is less than 1 cm.
Problems 12.3.
1. A man drops out of a plane at 25,000 feet of altitude and immediately opens his parachute. For
this man and parachute the deceleration due to air resistance is proportional to 4v where v is his
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velocity. How far has he fallen in one second? How long does it take for him to hit the ground,
and at what velocity does he hit the ground?
2. a) Let a mass m hang from a spring of spring constant k. Suppose that it is set in motion.
Show that, throughout the motion, mv 2 + kx2 is constant, where x represents displacement from
equilibrium, and v is velocity.
b) Suppose that k = 4 dynes/cm and m = 10 g, and the spring is already in motion At a particular
instant the spring is located 10 cm. from equilibrium and traveling at velocity 8 cm/sec. For this
motion, what are the maximum velocity and maximum displacement of the mass?
3. The above configuration is put in a viscous fluid which exerts a retardant force proportional to
the velocity, with constant of proportionality q = 12. Find the equation of motion of the mass,
given that at time t = 0 it is at x = 0 and its velocity is 4.8 cm/sec. What is the maximum
displacement of the mass?
4. A 10 g mass is hung on a spring with spring constant 0.8 dynes/cm. The mass is extended 40
cm beyond equilibrium and then released. What is the maximum speed the mass attains?
5. The system described in problem 4 is now inserted in a fluid with constant of viscosity q = 4.
The mass is extended 10 cm and released. Find the equation of motion.
12.4 The Inhomogeneous Equation
We return now to the general second order equation with a nonzero right hand side;
(12.14)
y 00 + ay 0 + by = g(x) ,
Proposition 12.6 . Suppose that y = yp (x) is a particular solution of the equation (12.14). Then
every solution is of the form y = yp + yh where yh is a solution of the homogeneous equation.
Example 12.12. Find the solution of the initial value problem:
y 00 + y = x + 2 ,
y(0) = 4 ,
y 0 (0) = 2 .
It is easy to see that yp (x) = x + 2 is a particular solution of this equation. Since the homogeneous
equation is y 00 + y = 0, the general solution is of the form
y(x) = x + 2 + A cos x + B sin x
To find A and B we use the initial conditions (at x = 0):
4 = 0 + 2 + A cos(0) ,
2 = 1 − A sin(0) + B cos(0)
giving us 4 = 2 + A, 2 = 1 + B, so A = 2, B = 1, and the solution is y = x + 2 + 2 cos x + sin x.
Example 12.13. Knowing that y = −(1/3) cos(2x) solves the differential equation
y 00 + y = cos(2x)
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find the solution with initial values y(π/2) = 1, y 0 (π/2) = 3 .
We know the solution has the form y = −(1/3) cos(2x) + A cos x + B sin x. Putting in the initial
values gives us 1 = 1/3 + B, 3 = −A, so the solution is y = −(1/3) cos(2x) − 3 cos x + (2/3) sin x.
In general, we may neither be given a particular solution, nor can we see one by inspection.
It is usually very difficult to find that first particular solution. However, here is one technique
(basically trial and error) which leads to a particular solution when the inhomogeneous function is
an elementary function.
Undetermined coefficients. To solve y 00 + ay 0 + by = g(x), try a function of the same form as
g(x). More precisely:
If g is a polynomial of degree n, try the general polynomial of degree n.
If g is an exponential times a polynomial of degree n, try the general exponential times a polynomial
of degree n.
If g is a cosine or a sine times a polynomial of degree n, try the general cosine and sine times a
polynomial of degree n.
The reason this works is that successive differentiation keeps us in the same form, so we end
up equating coefficients and solving for them. There is one caution: if g is a solution of the
homogeneous equation, this will fail In this case we replace the phrase “polynomial of degree n”
by “polynomial of degree n or more”. Let us illustrate
Example 12.14. Find a particular solution of y 00 + 5y 0 − y = x2 − 3x + 4.
We try y = ax2 + bx + c. First we calculate the first and second derivative: y 0 = 2ax + b, y 00 = 2a.
Substituting these in the given equation we obtain
2a + 5(2ax + b) − (ax2 + bx + c) = x2 − 3x + 4
This simplifies to
−ax2 + (10a − b)x + 2a + 5b − c = x2 − 3x + 4
We now equate coefficients:
−a = 1 ,
10a − b = −3 ,
2a + 5b − c = 4
giving the solutions a = −1, b = −7, c = −41, so the answer is
y = −x2 − 7x − 41 .
Example 12.15. Find a particular solution of y 00 + y 0 − y = xex .
Try y = (ax + b)ex . Differentiating, y 0 = (ax + a + b)ex , y 00 = (ax + 2a + b)ex . Substituting in the
given equation leads to
(ax + 2a + b)ex + (ax + a + b)ex − (ax + b)ex = xex ,
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reducing to the equations a = 1, 3a + b = 0. Thus the answer is
y = (x − 3)ex
Example 12.16. Find a particular solution of y 00 + y = cos x.
Since cos x satisfies the homogeneous equation, we must try y = ax sin x + bx cos x. Then y 00 =
2a cos x − 2b sin x − ax sin x − bx cos x, and we obtain the equation
2a cos x − 2b sin x = cos x
so a = 1/2, and the solution is y = (1/2)x sin x.
Problems 12.4
1. Solve y 00 + 2y 0 + y = x with the initial values y(0) = 0, y 0 (0) = 0.
2. Find the general solution of y 00 + 2y 0 + y = sin x.
3. Find the general solution of y 00 − 4y = sin(2x).
4. Find the general solution of y 00 + 4y = sin(2x).
5. A crystal glass consists of cells in a crystalline shape which oscillate at a natural frequency, so
the motion is governed by a differential equation x00 + ω02 x = 0 where 2π/ω0 is the frequency. If
the ambient air is vibrating at a frequency of ω/2π (due to a monotonal sound, perhaps), then
the motion of a crystal is modified by the force of the air in motion so as to be governed by the
inhomogeneous equation
x00 + ω02 x = A cos ωt
Find a particular solution of this differential equation. What happens as ω approaches ω0 ? (This
phenomenon is called resonance.)
240